# Document 13351933

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10.3 Convergence Tests All Series
23 If Za, does not converge show that Zla,l does not
converge.
24 Find conditions which guarantee that a,
+ a, - a3 +
a, + a5 - a, + -.- will converge (negative term follows two
positive terms).
25 If the terms of In 2 = 1- 4 + 3 - f
+ --.are rearranged into
1- 3 - 6 + 4 - - &amp; + --.,show that this series now adds to
4 In 2. (Combine each positive term with the following negative term.)
34 Verify the Schwarz inequality (Ca, bJ2 &lt; (CaZ)(Zbi) if
a,
= (4)&quot; and
b, = (4)&quot;.
a2
35 Under what condition does ?(a,+, -an) converge and
what is its sum?
36 For a conditionally convergent series, explain how the
terms could be rearranged so that the sum is + co. All terms
must eventually be included, even negative terms.
37 Describe the terms in the product (1
26 Show that the series 1 + 4 to 4 In 2.
+4 +4 - +
+ 4 + f + .--)(I+ 4 +
38 True or false:
+4-&amp;+
27 What is the sum of 1 +*-\$+*-f
1
28 Combine 1 + - - . + - - l n n + y
n
to prove 1 + * + 4 - 3 - \$ - &amp; +
converges
4 + ---)and find their sum.
-.a?
and 1 - \$ + ~ - . - + l n 2
= ln2.
29 (a) Prove that this alternating series converges:
(a) Every alternating series converges.
(b) Za, converges conditionally if Z la,l diverges.
(c) A convergent series with positive terms is absolutely
convergent.
(d) If Can and Cb, both converge, so does C(a, + b,).
39 Every number x between 0 and 2 equals 1 + 4 +4+ ..with suitable terms deleted. Why?
+
(b) Show that its sum is Euler's constant y.
40 Every numbers between -1 and 1 equals f f\$ f\$ f --.
with a suitable choice of signs. (Add 1 = 4+ f + 4 + --.to get
Problem 39.) Which signs give s = - 1 and s = 0 and s = i?
30 Prove that this series converges. Its sum is 4 2 .
41 Show that no choice of signs will make
equal to zero.
1 1
+ -.- .--.Compute
2! 4!
cos 1 to five correct decimals (how many terms?).
31 The cosine of 8 = 1 radian is 1 - -
It3
32 The sine of 8 = 7~ radians is n - -
sin 7~
715
+-
ma..
3! 5!
to eight correct decimals (how many terms?).
Compute
33 If Xai and Zbi are convergent show that Za,b, is absolutely convergent.
Hint: (a fb)2 2 0 yields 2)abJ&lt; a2 + b2.
+ 4+ 4\$ &amp; + .--
42 The sums in Problem 41 form a Cantor set centered at
zero. What is the smallest positive number in the set? Choose
signs to show that 4 is in the set.
&quot;43 Show that the tangent of 0 = q(n - 1) is sin 1/(1 - cos 1).
This is the imaginary part of s = - ln(1 - ei). From
s = Z ein/ndeduce the remarkable sum C (sin n)/n = q(7~- 1).
-
44 Suppose Can converges and 1x1 &lt; 1. Show that Ca,xn
converges absolutely.
10.4 The Taylor Series for ex,sin x, and cos x
This section goes back from numbers to functions. Instead of Xu, = s it deals with
Xanxn=f(x). The sum is a function of x. The geometric series has all a, = 1 (including
a,, the constant term) and its sum isf (x) = 1/(1- x). The derivatives of 1 x x2 --.
match the derivatives off. Now we choose the an differently, to match a different
function.
The new function is ex. All its derivatives are ex. At x = 0, this function and its
derivatives equal 1. To match these l's, we move factorials into the denominators.
+ + +
10 Infinite Series
Term by term the series is
xn/n!has the correct nth derivative (= 1). From the derivatives at x = 0, we have built
back the function! At x = 1 the right side is 1 + 1 + 4+ &amp; + .-.and the left side is e =
2.71828 .... At x = - 1 the series gives 1- 1 + f - + -, which is e-'.
The same term-by-term idea works for differential equations, as follows.
4
EXAMPLE 1 Solve dyldx =
- y starting from y = 1 at x = 0.
Solution The zeroth derivative at x = 0 is thefunction itseg y = 1. Then the equation
y' = - y gives y' = - 1 and y&quot; = - y' = + 1. The alternating derivatives
1, -1, 1, - 1, ... are matched by the alternating series for e-&quot;:
y = 1 - x + t x 2 - i x 3 + ... - e - X (the correct solution to y' = - y).
EXAMPLE 2 Solve d'y/dx2 = - y starting from y = 1 and y' = 0 (the answer is cos x).
Solution The equation gives y&quot; = - 1(again at x = 0). The derivativeof the equation
= - y&quot; =
1. The even derivatives are alternately + 1
gives y'&quot; = - y1= 0. Then
and - 1, the odd derivatives are zero. This is matched by a series of even powers,
which constructs cos x:
1
1
1
y = 1 - -X2 +
- -X6 + ... = cos X.
2!
4!
6!
+
The first terms 1 - \$x2 came earlier in the book. Now we have the whole alternating
series. It converges absolutely for all x, by comparison with the series for ex (odd
powers are dropped). The partial sums in Figure 10.4reach further and further before
they lose touch with cos x.
Fig. 10.4 The partial sums 1 - x2/2 + x4/24 - --.of the cosine series.
If we wanted plus signs instead of plus-minus, we could average ex and e-&quot;. The
differential equation for cosh x is d2y/dx2= + y, to give plus signs:
1
-(ex
2
1
1
1
+ e-&quot;) = 1 + -x2
+ -x4
+ -x6
+
2!
4!
6!
(which is cosh x).
W O R SERIES
The idea of matching derivatives by powers is becoming central to this chapter. The
derivatives are given at a basepoint (say x = 0). They are numbers f (O),f '(O), .... The
derivative [email protected])(O)will be the nth derivative of anxn,if we choose a, to be f(&quot;)(O)/n!
387
0.4A The Taylor Series for eX, sin ;x and cos x
Then the series I anx&quot; has the same derivatives at the basepoint as f(x):
10K The Taylor series that matches f(x) and all its derivatives at x = 0 is
f(0) + f'(0)x +
11f&quot; 0, 222.......
1, +
+ 6 '(0)x
+
+
= n=O fn)(0)...
n! x.
The first terms give the linear and quadratic approximations that we know well. The
x3 term was mentioned earlier (but not used). Now we have all the terms-an &quot;infinite
approximation&quot; that is intended to equalf(x).
Two things are needed. First, the series must converge. Second, the function must
do what the series predicts, away from x = 0. Those are true for ex and cos x and
sin x; the series equals the function. We proceed on that basis.
The Taylor series with special basepoint x = 0 is also called the &quot;Maclaurin series.&quot;
EXAMPLE 3
Find the Taylor series for f(x) = sin x around x = 0.
Solution The numbers f(&quot;)(0 ) are the values of f= sin x, f' = cos x, f&quot; = - sin x,...
All even derivatives are zero. To find
at x = 0. Those values are 0, 1,0, -1, 0, 1,....
the coefficients in the Taylor series, divide by the factorials:
(2)
X5 sin x= x- ix 3&plusmn; +
EXAMPLE 4 Find the Taylor series forf(x) = (1 + x)5 around x = 0.
Solution This function starts at f(0) = 1. Its derivative is 5(1 + x)4 , so f'(0) = 5. The
second derivative is 5 . 4 .(1 + x)3 , so f &quot;(0)= 5 4. The next three derivatives are
5 . 4 *3, 5 *4 . 3 *2, 5 . 4 . 3 *2 * 1. After that all derivatives are zero. Therefore the Taylor
series stops after the x s term:
5.4
5.4.3
5-4-3-2 &quot; 1
5-4-3-2
5
x2
+
x4
.
(3)
2!
3!
4!
5!
You may recognize 1, 5, 10, 10, 5, 1. They are the binomialcoefficients, which appear
in Pascal's triangle (Section 2.2). By matching derivatives, we see why 0!, 1!, 2!, ... are
needed in the denominators.
1+ 5x+
There is no doubt that x = 0 is the nicest basepoint. But Taylor series can be constructed around other points x = a. The principle is the same-match derivatives by
powers-but now the powers to use are (x - a)&quot;. The derivatives f(&quot;'(a) are computed
at the new basepoint x = a.
The Taylor series begins with f(a) +f'(a)(x - a). This is the tangent approximation
at x = a. The whole &quot;infinite approximation&quot; is centered at a-at that point it has
the same derivatives as f(x).
10L The Taylor series for f(x) around the basepoint x = a is
+
f(x) = f(a) + f(a)(x - a)+
(•
2n
(a)(x
+(a).. ...f
a) +
=
=O
n
&quot;,
(x- a)
(4)
EXAMPLE 5 Find the Taylor series forf(x) = (1+ x)5 around x = a = 1.
Solution At x = 1, the function is (1+ 1)' = 32. Its first derivative 5(1 + x)4 is
5-16 = 80. We compute the nth derivative, divide by n!, and multiply by (x - 1)&quot;:
32 + 80(x - 1)+ 80(x - 1)2 + 40(x - 1)3 + 10(x - 1)4 + (x - 1)5.
I 0 infinite Series
+
That Taylor series (which stops at n = 5) should agree with (1 x)'. It does. We could
rewrite 1 x as 2 (x - I), and take its fifth power directly. Then 32, 16, 8,4,2, 1 will
multiply the usual coefficients 1,.5, 10, 10, 5, 1 to give our Taylor coefficients
32, 80, 80,40, 10, 1. The series stops as it will stop for any polynomial-because the
high derivatives are zero.
+
+
EXAMPLE 6 Find the Taylor series for f(x) = ex around the basepoint x = 1.
Solution At x = 1 the function and all its derivatives equal e. Therefore the
Taylor series has that constant factor (note the powers of x - 1, not x):
DEFINING THE FUNCTION BY ITS SERIES
Usually, we define sin x and cos x from the sides of a triangle. But we could start
instead with the series. Define sin x by equation (2). The logic goes backward, but it
is still correct:
First, prove that the series converges.
Second, prove properties like (sin x)' = cos x.
Third, connect the definitions by series to the sides of a triangle.
We don't plan to do all this. The usual definition was good enough. But note first:
There is no problem with convergence. The series for sin x and cos x and ex all have
terms fxn/n!. The factorials make the series converge for all x. The general rule for
ex times eYcan be based on the series. Equation (6) is typical: e is multiplied by
powers of (x - 1). Those powers add to ex-'. So the series proves that ex = eex-'.
= ex+Y:
That is just one example of the multiplication (ex)(eY)
Term by term, multiplication gives the series for ex+Y.Term by term, differentiating
the series for ex gives ex. Term by term, the derivative of sin x is cos x:
We don't need the famous limit (sin x)/x -, 1, by which geometry gave us the derivative. The identities of trigonometry become identities of infinite series. We could even
define n as the first positive x at which x - i x 3 + .--equals zero. But it is certainly
not obvious that this sine series returns to zero-much less that the point of return
is near 3.14.
The function that will be dejined by injnite series is eie. This is the exponential of
the imaginary number i0 (a multiple of i =
The result eiBis a complex number,
and our goal is to identify it. (We will be confirming Section 9.4.) The technique is to
treat i0 like all other numbers, real or complex, and simply put it into the series:
fl).
1
DEFINITION eie is the sum of 1 + (i0) + -(i0)2
2!
1
+ -(i0)3
+ -.-.
3!
(9)
Now use iZ= - 1. The even powers are i4= + 1, i6 = - 1, i8 = + 1, .... We are
just multiplying - 1 by - 1 to get 1. The odd powers are i 3 = - i, is = + i, .... There-
10.4 The Taylor Series for d, sin x, and cos x
fore eiBsplits into a real part (with no i's) and an imaginary part (multiplying i):
You recognize those series. They are cos 8 and sin 8. Therefore:
y = r sin 8
eie = cos 0 + i sine
I
Euler's formula is eie= cos 8 + i sin 8. Note that e2&quot;' = 1.
The real part is x = cos 8 and the imaginary part is y = sin 8. Those coordinates pick
out the point eiBin the &quot;complex plane.&quot; Its distance from the origin (0,O) is r = 1,
because (cos 8)2+ (sin 8)2= 1. Its angle is 8, as shown in Figure 10.5. The number
-1 is ei&quot;, at the distance r = l and the angle n. It is on the real axis to the left of
zero. If eiBis multiplied by r = 2 or r = 3 or any r 2 0, the result is a complex number
at a distance r from the origin:
x=r'cose 1 Rg. 10.5
Complex numbers: reiB= r(cos 8 + i sin 8) = r cos 8 + ir sin 8 = x + iy.
With eie, a negative number has a logarithm. The logarithm of - 1 is imaginary
(it is in, since ei&quot; = - 1). A negative number also has fractional powers. The fourth
root of - 1 is (- l)'I4 = einI4.More important for calculus: The derivative of x5I4 is
\$x1I4. That sounds old and familiar, but at x = - 1 it was never allowed.
Complex numbers tie up the loose ends left by the limitations of real numbers.
+
The formula eie= cos 8 i sin 8 has been called &quot;one of the greatest mysteries of
undergraduate mathematics.&quot; Writers have used desperate methods to avoid infinite
series. That proof in (10) may be the clearest (I remember sending it to a prisoner
studying calculus) but here is a way to start from d/dx(eix)= ieix.
A diferent proof of Euler'sformula Any complex number is eix= r(cos 8 + i sin 8)
for some r and 8. Take the x derivative of both sides, and substitute for ieix:
(cos 8 + i sin B)dr/dx + r(- sin 8 + i cos B)d8/dx = ir(cos 8 + i sin 9).
Comparing the real parts and also the imaginary parts, we need drldx = 0 and
d8/dx = 1. The starting values r = 1 and 8 = 0 are known from eiO= 1. Therefore r is
always 1 and 8 is x. Substituting into the first sentence of the proof, we have Euler's
formula eie= l(cos 8 + i sin 8).
- - -
-
-
-
-
-
The a series is chosen to match f(x) and all its b
at the basepoint. Around x = 0 the series begins with
f(0) + c x + d x 2 . The coefficient of x n is e . For
f
. For f ( x )= cos x the series is
f ( x ) = ex this series is
g . For f ( x ) = sin x the series is
h . If the signs were
all positive in those series, the functions would be cosh x and
I . Addition gives cosh x + sinh x =
I .
In the Taylor series for f ( x ) around x = a, the coefficient of
( x - a)&quot; is b, = k . Then b,(x - a)&quot; has the same I as
f a t the basepoint. In the example f ( x ) = x2, the Taylor coefficients are bo = m , b , = n , b2 = 0 . The series
bo + b , ( x - a ) + b2(x agrees with the original p .
The series for ex around x = a has bn = q . Then the
Taylor series reproduces the identity ex = ( r
8
).
We define ex, sin x, cos x, and also eie by their series. The
derivative d/dx(l + x + i x + -.-)= 1 + x + --. translates to
t
.Thederivativeofl-+xZ+-.-is u . u s i n g i 2 =
+ - - - splits into eie= v . Its
- 1 the series 1 + ie +
square gives e2&quot; = w . Its reciprocal is e-&quot; = x .
Multiplying by r gives reie= Y + i
,which connects
the polar and rectangular forms of a A number. The
.
logarithm of eie is
1 Write down the series for e2&quot;and compute all derivatives
at x = 0. Give a series of numbers that adds to e2.
2 Write down the series for sin 2x and check the third
derivative at x = 0. Give a series of numbers that adds to
sin 211 = 0.
390
10 Infinite Series
In 3-8 find the derivatives off (x) at x = 0 and the Taylor series
(powers of x) with those derivatives.
*34 For x &lt; 0 the derivative of xn is still nxn-':
+
3 f(x) = eix
4 f(x) = 1/(1 x)
5 f(x) = 1/(1 - 2x)
6 f(x) = cosh x
What is dlxlldx? Rewrite this answer as nxn- '.
&amp;
Problems 9-14 solve differential equations by series.
9 From the equation dyldx = y - 2 find all the derivatives
of y at x = 0 starting from y(0) = 1. Construct the infinite
series for y, identify it as a known function, and verify that
the function satisfies y' = y - 2.
35 Why doesn't f(x) =
have a Taylor series around x =
O? Find the first two terms around x = 1.
36 Find the Taylor series for 2&quot; around x = 0.
In 37-44 find the first three terms of the Taylor series around
x = 0.
10 Differentiate the equation y' = cy + s (c and s constant)
to find all derivatives of y at x = 0. If the starting value is
yo = 0, construct the Taylor series for y and identify it with
the solution of y' = cy + s in Section 6.3.
37 f(x) = tan-'x
38 f (x) = sin - 'x
39 f(x) =tan x
40 f (x) = ln(cos x)
11 Find the infinite series that solves y&quot; = - y starting from
y=O and y'= 1 at x = 0 .
43 f (x) = cos2x
44 f (x) = sec2x
12 Find the infinite series that solves y' = y starting from y =
45 From eie= cos 6 + i sin 8 and e-&quot; = cos 6 - i sin 6, add
and subtract to find cos 8 and sin 8.
1 at x = 3 (use powers of x - 3). Identify y as a known
function.
46 Does (eiB)2equal cos28 i sin26 or cos O2
13 Find the infinite series (powers of x) that solves y&quot; =
47 Find the real and imaginary parts and the 99th power of
ei&quot;, ei&quot;/2 ei&quot;/4' and e-'&quot;I6.
2y' - y starting from y = 0 and y'
= 1 at
x = 0.
14 Solve y&quot; = y by a series with y = 1 and y' = 0 at x = 0 and
identify y as a known function.
15 Find the Taylor series for f(x) = (1 + x ) ~
around x = a =
0 and around x = a = 1 (powers of x - 1). Check that both
series add to (1 + x ) ~ .
16 Find all derivatives of f(x) = x3 at x = a and write out the
Taylor series around that point. Verify that it adds to x3.
17 What is the series for (1 - x)' with basepoint a = l ?
18 Write down the Taylor series for f = cos x around x = 21t
and also for f = cos (x - 21t) around x = 0.
In 19-24 compute the derivatives off and its Taylor series
around x = 1.
22 f(x) = x4
21 f(x) =In x
In 25-33 write down the first three nonzero terms of the Taylor
series around x = 0, from the series for ex, cos x, and sin x.
25 xe2X
sin x
28 -
sin x
dx
32 bx =
9
48 The three cube roots of 1 are 1, e2&quot;'I3, e4&quot;'I3.
(a) Find the real and imaginary parts of e2&quot;'I3.
(b) Explain why (e2&quot;i13)3= 1.
(c) Check this statement in rectangular coordinates.
49 The cube roots of -1 = ei&quot; are ei&quot;I3 and
and
. Find their sum and their product.
fi
50 Find the squares of 2eid3= 1 +
i and 4ei'I4
2 f i + i2&amp; in both polar and rectangular coordinates.
=
51 Multiply eis= cos s + i sin s times eit= cos t + i sin t to
find formulas for cos(s + t) and sin(s + t).
52 Multiply eis times e-&quot; to find formulas for cos(s - t) and
sin(s - t).
53 Find the logarithm of i. Then find another logarithm of i.
(What can you add to the exponent of elniwithout changing
the result?)
54 (Proof that e is irrational) If e = p/q then
30 sin x2
33 ex cos x
55 Solve dyldx = y by infinite series starting from y = 2 at
x = 0.
X
31 ex'
+ i sin 02?
would be an integer. (Why?) The number in brackets-the
distance from the alternating series to its sum lle-is less
than the last term which is llp! Deduce that IN1 c 1 and reach
a contradiction, which proves that e cannot equal plq.
27 (1 - cos x)/x2
26 cos
+
391
10.5 Power Series
10.5 Power Series
This section studies the properties of a power series. When the basepoint is zero, the
powers are x&quot;. The series is anx&quot;. When the basepoint is x = a, the powers are
(x - a)&quot;. We want to know when and where (and how quickly) the series converges
to the underlying function. For ex and cos x and sin x there is convergence for all
x-but that is certainly not true for 1/(1 - x). The convergence is best when the
function is smooth.
First I emphasize that power series are not the only series. For many applications
they are not the best choice. An alternative is a sum of sines, f(x) = b,sin nx. That
is a &quot;Fouriersine series&quot;, which treats all x's equally instead of picking on a basepoint.
A Fourier series allows jumps and corners in the graph-it takes the rough with the
smooth. By contrast a power series is terrific near its basepoint, and gets worse as
you move away. The Taylor coefficients an are totally determined at the basepoint-where all derivatives are computed. Remember the rule for Taylor series:
an = (nth derivative at the basepoint)/n! =f(&quot;)(a)/n!
A remarkable fact is the convergence in a symmetric interval around x = a.
(1)
40M A power series Zax&quot; either converges for all x, or it converges only at
the basepoint x =0, or else it has a radius of convergence r:
Ylax&quot; converges absolutely if lxl
&lt; r and diverges ifxi &gt; r.
The series Ix&quot;/n! converges for all x (the sum is ex). The series In!x&quot; converges for
no x (except x = 0). The geometric series Ex&quot; converges absolutely for Ixl &lt; 1 and
diverges for Ixl
&gt; 1. Its radius of convergence is r = 1. Note that its sum 1/(1 - x) is
perfectly good for Ix&gt; 1-the function is all right but the series has given up. If
something goes wrong at the distance r, a power series can't get past that point.
When the basepoint is x = a, the interval of convergence shifts over to Ix - al &lt; r.
The series converges for x between a - r and a + r (symmetric around a). We cannot
say in advance whether the endpoints a + r give divergence or convergence (absolute
or conditional). Inside the interval, an easy comparison test will now prove convergence.
PROOF OF 10M Suppose Ya,X&quot; converges at a particular point X. The proof will
show that Ia,x&quot; converges when Ixi
is less than the number IXI. Thus convergence
at X gives convergence at all closer points x (I mean closer to the basepoint 0). Proof:
Since I anX&quot; converges, its terms approach zero. Eventually lanX&quot;I &lt; 1 and then
Iaxnl = IaX&quot;I Ix/lX
&lt; Ix/XI&quot;.
Our series I a,x&quot; is absolutely convergent by comparison with the geometric series
for Ix/XI, which converges since Ix/XI &lt; 1.
EXAMPLE 1 The series Inx&quot;/4&quot; has radius of convergence r = 4.
The ratio test and root test are best for power series. The ratios between terms
approach x/4 (and so does the nth root of nx&quot;/4&quot;):
(n + 1)x&quot;+' Inx&quot;
/4
4n+1
x n+ 1
x
n approaches L = 4
-= 4
The ratio test gives convergence if L &lt; 1, which means Ixl &lt; 4.
392
10 Infinite Series
X3
EXAMPLE 2 The sine series x -
5
+ 3! 5!
-
has r = co (it converges everywhere).
The ratio of xn+ 2/(n + 2)! to x&quot;/n! is x 2/(n + 2)(n + 1). This approaches L = 0.
EXAMPLE 3
The series I(x - 5)&quot;/n 2 has radius r = 1 around its basepoint a = 5.
The ratios between terms approach L = x - 5. (The fractions n2/(n + 1)2 go toward
1.) There is absolute convergence if Ix - 51 &lt; 1. This is the interval 4 &lt; x &lt; 6, symmetric around the basepoint. This series happens to converge at the endpoints 4 and 6,
because of the factor 1/n2 . That factor decides the delicate question-convergence at
the endpoints-but all powers of n give the same interval of convergence 4 &lt; x &lt; 6.
CONVERGENCE TO THE FUNCTION: REMAINDER TERM AND RADIUS r
Remember that a Taylor series starts with a function f(x). The derivatives at the
basepoint produce the series. Suppose the series converges: Does it converge to
the function? This is a question about the remainderR,(x) =f(x) - sn(x), which is the
difference between f and the partial sum s, = ao + &quot; + an(x - a)&quot;. The remainder Rn
is the error if we stop the series, ending with the nth derivative term a,(x - a)&quot;.
10N Suppose f has an (n + 1)st derivative from the basepoint a out to x. Then
for some point c in between (position not known) the remainder at x equals
R,(x)= f(x) - s,(x) = f(&quot;+ 1)(c)(x - a)&quot; +'/(n + 1)!
(2)
The error in stopping at the nth derivative is controlled by the (n + 1)st derivative.
You will guess, correctly, that the unknown point c comes from the Mean Value
Theorem. For n = 1 the proof is at the end of Section 3.8. That was the error e(x) in
linear approximation:
R1 (x) = f(x) - f(a) - f'(a)(x - a) =
f &quot;(c)(x - a)2.
&quot;
For every n, the proof compares Rn with (x - a) +1. Their (n + 1)st derivatives are
and (n + 1)! The generalized Mean Value Theorem says that the ratio of R, to
f(n+ ')
(x - a)&quot; + equals the ratio of those derivatives, at the right point c. That is equation
(2). The details can stay in Section 3.8 and Problem 23, because the main point is
what we want. The error is exactly like the next term a,+ (x - a)&quot; +1, except that the
(n + 1)st derivative is at c instead of the basepoint a.
EXAMPLE 4
Whenf is ex, the (n + 1)st derivative is ex. Therefore the error is
Rn=ex - 1 +x+ ... + -
&quot;
= ec
n! = (n+ 1)!*
(3)
At x = 1 and n = 2, the error is e - (1 + 1 + ) .218. The right side is ec/6. The
unknown point is c = In(.218 - 6) = .27. Thus c lies between the basepoint a = 0 and
the error point x = 1, as required. The series converges to the function, because
R, - 0.
In practice, n is the number of derivatives to be calculated. We may aim for an
error IRn, below 10- 6. Unfortunately, the high derivative in formula (2) is awkward
to estimate (except for ex). And high derivatives in formula (1) are difficult to compute.
Most real calculations use only afew terms of a Taylor series. For more accuracy we
move the basepoint closer, or switch to another series.
10.5 Power Serles
There is a direct connection between the function and the convergence radius r.
also ends
the convergence interval for the series. Another hint comes for f = llx, if we expand
around x = a = 1:
A hint came for f(x) = 1/(1- x). The function blows up at x = 1-which
This geometric series converges for 11 - XI&lt; 1. Convergence stops at the end point
x = 0-exactly where llx blows up. The failure of the function stops the convergence
of the series. But note that 1/(1+ x2), which never seems to fail, also has convergence
1/(1+ x2) = 1 - x2 + x4 - x6 +
converges only for 1x1 &lt; 1.
When you see the reason, you will know why r is a &quot;radius.&quot; There is a circle, and
the function fails at the edge of the circle. The circle contains complex numbers as
well as real numbers. The imaginary points i and - i are at the edge of the circle.
The function fails at those points because l/(l + i2)= co.
Complex numbers are pulling the strings, out of sight. The circle of convergence
reaches out to the nearest &quot;singularity&quot; of f(x), real or imaginary or complex. For
1/(1+ x2), the singularities at i and - i make r = 1. If we expand around a = 3, the
distance to i and - i is r = ,/%. If we change to in (1 + x), which blows up at
X = - 1, the radius of convergence of x - \$x2 + gx3is r = 1.
.-•
a=0
ln(1
1/(1 + x 2 ) =
-
+ x ) and ( 1 + x
) ~
also at -i
Fig. 10.6 Convergence radius r is distance from basepoint a to nearest singularity.
THE BINOMIAL SERIES
We close this chapter with one more series. It is the Taylor series for (1 + x ) ~around
,
the basepoint x = 0.A typical power is p = 3, where we want the terms in
JK=
1+ f x + a 2 x 2 +
em-.
The slow way is to square both sides, which gives 1 + x + (2a2+ \$)x2 on the right.
Since 1 + x is on the left, a2 = - &amp; is needed to remove the x2 term. Eventually a,
can be found. The fast way is to match the derivatives off = (1 + x)'I2:
At x = 0 those derivatives are 4,- i,
8. Dividing by I!, 2!, 3! gives
These are the binomial coeficients when the power is p = 4.
Notice the difference from the binomials in Chapter 2. For those, the power p was
~ 1 2x + x2 stopped at x2. The coefficients for
a positive integer. The series (1 + x ) =
p = 2 were 1,2, 1,0,0,0, .... For fractional p or negative p those later coefficients are
not zero, and we find them from the derivatives of (1 + x ) ~ :
+
Dividing by O!, I!, 2!, ..., n! at x = 0, the binomial coefficients are
For p = n that last binomial coefficient is n!/n! = 1. It gives the final xn at the end of
(1 + x)&quot;. For other values of p, the binomial series never stops. It convergesfor 1x1&lt; 1:
When p = 1,2, 3, ... the binomial coeflcient p!/n!(n - p)! counts the number of ways
to select a group of n friends out of a group of p friends. If you have 20 friends, you
can choose 2 of them in (20)(19)/2= 190 ways.
Suppose p is not a positive integer. What goes wrong with (1 + x ) ~ to
, stop the
~
up.
convergence at 1x1 = l? The failure is at x = - 1. If p is negative, (1 + x ) blows
the
, higher derivatives blow up. Only for a positive
If p is positive, as in ,/=
integer p = n does the convergence radius move out to r = GO. In that case the series
for (1 + x)&quot; stops at xn, and f never fails.
A power series is a function in a new form. It is not a simple form, but sometimes
it is the only form. To compute f we have to sum the series. To square f we have to
multiply series. But the operations of calculus-derivative and integral-are easier.
That explains why power series help to solve differential equations, which are a rich
source of new functions. (Numerically the series are not always so good.) I should
have said that the derivative and integral are easy for each separate term anxn-and
fortunately the convergence radius of the whole series is not changed.
Iff (x) = Xanxnhas convergence radius r, so do its derivative and its integral:
df/dx = C nanxn-' and
If (x)dx= Z anxn '/(n + 1) also converge for 1x1 &lt; r.
+
EXAMPLE 5 The series for 1/(1- x) and its derivative 1/(1- x ) ~and its integral
- ln(1 - x) all have r = 1 (because they all have trouble at x = 1). The series are Exn
and Enxn-' and Cxn+'/(n + 1).
EXAMPLE 6 We can integrate ex' (previously impossible) by integrating every term
in its series:
This always converges (r = a).The derivative of ex' was never a problem.
395
10.5 Power Series
10.5 EXERCISES
If 1x1 &lt; IX I and Ca,Xn converges, then the series C.a,xn also
a
. There is convergence in a b interval around the
c
. For C(2x)&quot; the convergence radius is r = d . For
C.xn/n! the radius is r = e . For C(x - 3)&quot; there is convergence for Ix - 31 &lt; f . Then x is between g and
h
.
Starting with f(x), its Taylor series Ca,xn has a, = i .
With basepoint a, the coefficient of (x - a)&quot; is i . The
error after the x n term is called the k R,(x). It is equal to
I
where the unknown point c is between m . Thus
the error is controlled by the n derivative.
The circle of convergence reaches out to the first point
where f(x) fails. For f = 4/(2 - x), that point is x = o .
Around the basepoint a = 5, the convergence radius would be
r = P . For sin x and cos x the radius is r = q .
The series for J l + x is the
series with p = f . Its
coefficients are a, = s . Its convergence radius is t .
Its square is the very short series 1 + x.
In 1-6 find the Taylor series forf (x) around x = 0 and its radius
of convergence r. At what point does f(x) blow up?
1 f(x) = 1/(1 - 4x)
2 f(x) = 1/(1 - 4x2)
3 f(x)= el-&quot;
4 f(x) = tan x (through x3)
5 f(x) = ln(e
+ x)
6 f(x) = 1/(1 +4x2)
15 From the series for (1 - cos x)/x2 find the limit as x + 0
faster than 1'H;pital's rule.
16 Construct a power series that converges for 0 &lt; x &lt; 27r.
17 If the cosine series stops before x8/8! show from (2) that
the remainder R7 is less than x8/8! Does this also follow
because the series is alternating?
18 If the sine series around x = 27r stops after the terms in
problem 10, estimate the remainder from equation (2).
'+
+
in
19 Estimate by (2) the remainder R, = xn+ xn+
the geometric series. Then compute R, exactly and find the
unknown point c for n = 2 and x = f .
+
20 For -ln(l - x ) = x + ~ x 2 + ~ x 3R3, useequation(2) to
show that R3 &lt; \$ at x = 3.
21 Find R, in Problem 20 and show that the series converges
to the function at x = f (prove that R, -+ 0).
22 By estimating R, prove that the Taylor series for ex around
x = 1 converges to ex as n -+ GO.
23 (Proof of the remainder formula when n = 2)
(a) At x = a find R,, R;, Ri, R;&quot;.
(b) At x = a evaluate g(x) = (x - a)3 and g', g&quot;, g&quot;'.
(c) What rule gives R2W - [email protected]) - R h ) ?
g(x) - d a )
g'(c 1)
Find the interval of convergence and the function in 7-10.
Ri(c2)- R;'(a) --Rll(c)
where are cl and c2 and c?
g&quot;'(c)
g&quot;(c2) - g&quot;(4
(e) Combine (a-b-c-d) into the remainder formula (2).
11 Write down the Taylor series for (ex- l)/x, based on the
series for ex. At x = 0 the function is 010. Evaluate the series
at x = 0. Check by l'H6pital's Rule on (ex- l)/x.
12 Write down the Taylor series for xex around x = 0. Integrate and substitute x = l to find the sum of l/n!(n + 2).
13 Iff (x) is an even function, so f (-x) =f (x), what can you
say about its Taylor coefficients in f = Ca,xn?
14 Puzzle out the sums of the following series:
(a)x+x2-x3+x4+x5--x6+-..
x4 x8
(b) 1 - - ...
4! 8!
+ + +
24 All derivatives off (x) = e- 'Ix2 are zero at x = 0, including
f(0) = 0. What is f(.l)? What is the Taylor series around
x = O? What is the radius of convergence? Where does the
series converge to f(x)? For x = 1 and n = 1 what is the
remainder estimate in (2)?
25 (a) Find the first three terms in the binomial series for
l/Ji7.
(b) Integrate to find the first three terms in the Taylor
series for sin - 'x.
26 Show that the binomial coefficients in I/,,/=
are a, = 1 3 5 (2n - 1)/2&quot;n!
= C.anxn
27 For p = - 1 and p = - 2 find nice formulas for the binomial coefficients.
28 Change the dummy variable and add lower limits to make
x w nxn- 1 - C&quot; (n
l)xn.
+
396
10 Infinite Series
29 In (1 - x)- ' = E x n the coefficient of x n is the number of
groups of n friends that can be formed from 1 friend (not
binomial-repetition
is allowed!). The coefficient is 1 and
there is only one group-the same friend n times.
(a) Describe all groups of n friends that can be formed
from 2 friends. (There are n + 1 groups.)
(b) How many groups of 5 friends can be formed from 3
friends?
30 (a) What is the coefficient of xn when 1 + x + x2 + . - multiplies 1 + x + x2 +
Write the first three terms.
(b) What is the coefficient of x5 in ( C X ~ ) ~ ?
40 (Composition of series) If f = a. + a,x + a2x2+ and
g = blx + b2x2+ ... find the 1, x, x2 coefficients of f(g(x)).
Test on f = 1/(1 + x), g = x/(l - x), with f(g(x)) = 1 - x.
- a .
41 (Multiplication of series) From the series for sin x and
1/(1 - x) find the first four terms for f = sin x/(l - x).
42 (Inversion of series) Iff = a l x + a2x2+ ... find coefficients
bl, b2 in g = b , x + b2x + ... SO that f (g(x))= x. Compute
bl,b2forf=ex-l,g=f-'=ln(l+x).
. a * ?
31 Show that the binomial series for ,/has integer
coefficients. (Note that xn changes to (4x)&quot;. These coefficients
are important in counting trees, paths, parentheses.. .)
32 In the series for l / ,/I
is (2n)! divided by (n!)2.
- 4x,
show that the coefficient of xn
Use the binomial series to compute 33-36 with error less than
1/1000.
33 (15)'14
34 (1001)'/~
35 (1.1)l.l
36
el/lOOO
37 From sec x = 1/[1- (1 - cos x)] find the Taylor series of
sec x up to x6. What is the radius of convergence r (distance
to blowup point)?
38 From sec2x = 1/[1 - sin2x] find the Taylor series up to
x2. Check by squaring the secant series in Problem 37. Check
by differentiating the tangent series in Problem 39.
39 (Division of series) Find tan x by long division of sin x/
COS x:
43 From the multiplication (sin x)(sin x) or the derivatives of
f(x) = sin2x find the first three terms of the series. Find the
first four terms for cos2x by an easy trick.
44 Somehow find the first six nonzero terms for f = (1 - x)/
(1 - x3).
45 Find four terms of the series for 1/J1the series to reach a geometric series.
46 Compute
power series.
Ji e P x 2dx
x. Then square
to 3 decimals by integrating the
47 Compute ji sin2t dt to 4 decimals by power series.
48 Show that Cxn/n converges at x = - 1, even though its
derivative Exn-' diverges. How can they have the same
49 Compute lim (sin x - tan x)/x3 from the series.
x+o
50 If the nth root of an approaches L &gt; 0, explain why C anxn
has convergence radius r = 1/L.
51 Find the convergence radius r around basepoints a = 0
and a = 1 from the blowup points of (1 + tan x)/(l + x2).
MIT OpenCourseWare
http://ocw.mit.edu
Resource: Calculus Online Textbook
Gilbert Strang
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